If Z = frac{(sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 | vedclass

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(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{

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$2$

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(C) Given $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$.Taking the modulus on both sides,we get $|Z| = \frac{|\sqrt{3} + i|^{3} |3i + 4|^{2}}{|8 + 6i|^{2}}$.Calculate the modulus of each complex number:$|\sqrt{3} + i| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$.$|3i + 4| = \sqrt{4^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$.$|8 + 6i| = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$.Substitute these values into the expression for $|Z|$:$|Z| = \frac{2^{3} 	imes 5^{2}}{10^{2}} = \frac{8 	imes 25}{100} = \frac{200}{100} = 2$.Thus,$|Z| = 2$.

If $Z = \frac{(\sqrt{3} + i)^{3}(3i + 4)^{2}}{(8 + 6i)^{2}}$,then $|Z|$ is equal to

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